静态成员初始化专用模板类 [英] static member initialization for specialized template class

问题描述

class A
{
};

template <typename A, int S>
class B
{
public:
        static int a[S];

        B()
        {
                a[0] = 0;
        }
};

template<> int B<A, 1>::a[1];

int main()
{
        B<A, 1> t;
        t;
}

它在GCC 4.1下编译，但不链接：

It compiles under GCC 4.1, but does not link:

static.cpp:(.text._ZN1BI1ALi1EEC1Ev[B<A, 1>::B()]+0x5): undefined reference to `B<A, 1>::a'

我希望如果可能，

I would prefer to keep initialisation specialised if it is possible, since the array holds some data specific to the type.

推荐答案

对于静态成员专用化，如果不初始化成员，它被作为一个专门化的声明，只是说哦，不要从主模板实例化成员，因为在别的地方有一个专门的定义。应该提到的是，定义应该出现在.cpp文件中（否则，你会得到相反的：多个定义），没有初始化器的声明仍然应该放在头文件中。

For static member specializations, if you don't initialize the member, it is taken as a specialization declaration, that just says "Oh, don't instantiate the member from the primary template, because there is a specialized definition somewhere else". It should be mentioned that the definition should appear in a .cpp file (otherwise, you will earn the opposite: multiple definitions), and the declaration without initializer should still be placed in the header file.

现在正确的语法确实如下，它应该不是出现在头文件中，而是 .cpp 文件

Now the correct syntax is indeed the following, and it should not appear in a header file, but in a .cpp file

template<> int B<A, 1>::a[1] = { };

以下内容应该仍然出现在头文件中：

The following should still appear in a header file:

template<> int B<A, 1>::a[1];

这将作为专业化声明。

因此，你不能专门化一个只有一个默认构造函数而且不可复制的成员，因为你需要这个语法：

From this, it follows that you can't specialize a member that only has a default constructor and is not copyable, because you would need this syntax:

// needs a copy constructor!
template<> Type Class<Arguments>::member = Type();

C ++ 0x修复此问题：

C++0x fixes this:

// doesn't anymore need a copy constructor
template<> Type Class<Arguments>::member{};

---

对于我们中的标准人，引号：

---

For the Standardese people among us, here are the quotes:

14.7.3 / 6 ：

如果模板，成员模板或类模板的成员是显式专用的，那么该专门化应在首次使用该专门化之前声明，这将导致隐式实例化发生，在每个翻译单元，其中发生这样的使用;不需要诊断。

If a template, a member template or the member of a class template is explicitly specialized then that specialization shall be declared before the first use of that specialization that would cause an implicit instantiation to take place, in every translation unit in which such a use occurs; no diagnostic is required.

14.7.3 / 15 ：

模板的静态数据成员的显式特化是定义，如果声明包含初始化器;否则，它是一个声明。 [注意：没有要求默认初始化的模板的静态数据成员的定义的语法。

An explicit specialization of a static data member of a template is a definition if the declaration includes an initializer; otherwise, it is a declaration. [Note: there is no syntax for the definition of a static data member of a template that requires default initialization.

template<> X Q<int>::x;

这是一个声明，不管X是否可以被默认初始化（8.5）。 ]

This is a declaration regardless of whether X can be default initialized (8.5). ]

3.2 / 3 ：

每个程序都应包含每个非内联函数或在该程序中使用的对象的一个​​定义;无需诊断。

Every program shall contain exactly one definition of every non-inline function or object that is used in that program; no diagnostic required.

3.2 / 5 ：

可以有一个类类型（第9节），枚举类型（7.2），带外部链接的内联函数（7.1.2），类模板），非静态函数模板（14.5.5），类模板（14.5.1.3）的静态数据成员，类模板（14.5.1.1）的成员函数或者未指定某些模板参数的模板特化[...]

There can be more than one definition of a class type (clause 9), enumeration type (7.2), inline function with external linkage (7.1.2), class template (clause 14), non-static function template (14.5.5), static data member of a class template (14.5.1.3), member function of a class template (14.5.1.1), or template specialization for which some template parameters are not specified (14.7, 14.5.4) in a program [...]

此限制为指定意味着我们允许进行以下操作，将其放入标题（因此可能具有该特化的多个定义）：

The restriction of this to "for which some template parameters are not specified" means that we are allowed to do the following, placing it into a header (thus possibly having multiple definitions of this specialization):

template<> template<typename T>
Type OuterClass<int>::InnerClass<T>::StaticMember = 0;

在您的情况下，您指定了所有参数，使其不包含在一个定义规则允许多个定义。

In your case, you have all parameters specified, making it not being covered by the one defintion rule for allowing multiple definitions.

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